close

Вход

Log in using OpenID

embedDownload
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Multiple choice questions
Answer all of the following questions. Read each question carefully. Fill
the correct bubble on your scantron sheet. Each question has exactly one
correct answer. All questions are worth the same amount of points.
1.
On a frictionless horizontal surface, three boxes are connected by
inextensible massless strings and are pulled by a force F as shown in
the figure. The masses of the boxes are proportional to their sizes.
Which string has to be the strongest so as not to break?
All 3 boxes have the same acceleration. The tension in one string is
the net force applied to the set of boxes and strings to the left of that
string, thus
TA = m A a
TB = ( m A + m B ) a
TC = (m A + m B + mC )a
Tc is the largest.
A. A
B. B
C. C
D. A, B and C must be equally strong
E. A and B must be equally strong.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
2.
The figure shows a top view of a ball on the end of a string traveling
counterclockwise in a circular path. The speed of the ball is constant.
If the string should break at the instant shown, the path that the ball
would follow is
A. 1
B. 2
After the string broke, the ball is free falling. Its path is a
parabola which viewed from the top is a straight line tangent to
the velocity of the ball when the string breaks.
C. 3
D. 4
E. impossible to tell from the given information
3. A 5-kg object undergoes a displacement ∆sr = −5iˆ + 2 ˆj . During the
r
displacement, a constant force F = 3iˆ + 8 ˆj acts on the object. All
r
F
on this
values are given in SI units. The work done by the force
object is (in Joules)
A.
B.
C.
D.
–15 J
16
-34
1
r
E.
6
r
W = F ⋅ ∆s = 3 × (−5) + 8 × 2 = 1
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
4.
A motor develops power as shown in the graph. The energy
expended by the motor in the time interval between t = 10 s and t = 30
s is
A.
B.
C.
D.
1.0 J
1.3 J
0.20 kJ
0.60 kJ
since P=dW/dt,
W =∫
30 s
10 s
E.
Pdt = 600 J (the area under the power curve between 10
and 30s).
0.70 kJ
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
5.
The shaded area in the figure represents a uniformly thick sheet of
metal. The center of mass of the sheet is closest to point
A. 1
B. 2
C. 3
By symmetry the CM is along the 213 line.
1 is the CM of the top and bottom parts of the sheet of metal.
Taking into account the right part, the CM moves to the right to
point 3.
D. 4
E. 5
6. You shoot an arrow with a mass of 0.54 kg from a bow. The bow
exerts a force of 125 N for 0.65s. The speed of the arrow as it leaves
the bow is
A. 0.10 km/s
B. 0.15 km/s
r
r
∆p = ∫ Fdt
v final =
F∆t 125 × 0.65
=
= 150 m / s
m
0.54
C. 0.23 km/s
D. 0.27 km/s
E. 0.30 km/s
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
7.
Two identical bodies of mass M move with equal speeds v. The
direction of their velocities is illustrated in the figure. The magnitude
of the linear momentum of the system is
In an (x,y) frame such the x axis is directed left to right and the y axis
bottom to top,
A. 2Mv
B. Mv
C. 4Mv
D.
r
r
ptotal = ∑ p = Mvxˆ + Mvyˆ
r
2
2
p total = (Mv ) + (Mv ) = 2 Mv
E.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
8.
A bicycle is moving at a speed v = 12.6 m/s. The wheels of the
bicycle roll without slipping. A small stone is stuck to one of the
tires. At the instant the stone is at point A in the figure, it comes free.
The velocity of the stone (magnitude and direction) relative to the
earth just after release is
A. 17.8 m/s at 45º above the horizontal, toward the front of the
bicycle.
Work in an (x,y) frame: x is directed to the left, and y is directed
up
The angular velocity of the wheel is
ω=
v
r
In the center of mass frame of the wheel the velocity of A is
r
v A / CM = ωryˆ = vyˆ
Relative to the Earth the velocity of the wheel is
r
r
r
v A / Earth = v A / CM + vCM / Earth = vyˆ + vxˆ
B.
C.
D.
E.
12.6 m/s at 45º above the horizontal, away from the bicycle.
12.6 m/s at 37º below the horizontal.
12.6 m/s straight up.
17.8 m/s at 45º above the horizontal, toward the back of the
bicycle.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
9.
A solid disk (Icm = ½ mR2) rolls without slipping up a plane a
distance s. The plane is inclined at an angle θ with the horizontal.
The disk has mass m, radius R, and an initial translational speed v.
How far does the disk travel up the incline?
A.
3v 2
4 g sin θ
Use the work energy theorem
2
1
1
1
11
3
 v 
KEi = mv 2 + Iω 2 = mv 2 +  mR 2   = mv 2
2
2
2
22
4
 R 
KE f = 0
Wnet = WWrDE = −mgs sin θ
KE f − KEi = Wnet ⇒ s =
B.
C.
v2
2 g sinθ
Rv
2g sinθ
D.
v2
2 g (sinθ − cosθ )
E.
v2
g sinθ
3v 2
4 g sin θ
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
10.
Assume that none of the objects is hollow.
If all of the objects illustrated in the figure have equal masses, which
object has the largest moment of inertia about the indicated axis?
Hint: for this question, you don't need to know the formulas giving
the moments of inertia of these particular solids. Just use the
definition of the moment of inertia: I = ∑ mi ri 2
A. ring
Imagine manufacturing all of these solids from the same block of
material. To maximize the moment of inertia, one needs to put
the parts making up the object as far as possible from the axis of
rotation. It is the ring that allows one to place all of the parts the
farthest from the axis of rotation.
B. cross
C. sphere
D. cube
E. rod
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Questions 11 through 17 all refer to the same problem.
For this problem, assume no friction. A mass m2 = 3.5 kg rests on a
horizontal table and is attached by strings to masses m1 = 1.5 kg and m3 =
2.5 kg as shown. The masses m1 and m3 hang freely.
g=9.8 m/s2
11. Initially the system of the 3 masses is held at rest by a hand that exerts
a normal force on m2 (the hand is not represented on the drawing
above).
r
r
T
If T1 is the tension by the string on m1 and 3 is the tension by the
string on m3, how do the magnitudes of the tensions compare when
the system is held at rest?
A.
B.
C.
D.
r
r
T1 > T3
r
r
T1 = T3
r
r
T1 < T3
When at rest, the tension force applied on m1 or m3 is equal to
the weight of m1 or m3. And we know m1<m3
Can't tell. There is not enough information.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
At t=0, the hand is removed from m2 and the 3 masses start moving from
rest.
12.
How do the magnitudes of the tensions
system is moving?
A.
B.
C.
r
r
T1 and T3 compare when the
r
r
T1 > T3
r
r
T1 = T3
r
r
T1 < T3
Since m1<m3, m1 moves up, m2 moves to the right and m3
moves down.
The system starts from rest, thus the acceleration of m2 is to the
right. Along an horizontal axis oriented left to right, Newton's
2nd law applied to m2 reads:
T3 − T1 = m2 a with a > 0
D.
Can't tell. There is not enough information.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
13. When the system is moving, what is the acceleration of mass m2 (in
m/s2)?
A. 0
B. 1.3
Apply Newton's 2nd law to m1 (along an axis directed up), to m2
(as done above), and to m3 (along an axis directed down)
T1 − m1 g = m1 a 

T2 − T1 = m2 a  ⇒ (m3 − m1 ) g = (m1 + m2 + m3 )a
m3 g − T2 = m3 a 
1
a=
× 9.8 = 1.3 m / s 2
7.5
C.
D.
E.
5.2
8.7
9.8
14. At instant t1, m3 has moved by 1m from its initial position. From the
initial position to the position at t1, how does the work done on m3 by
the weight of m3, Wwr 3 E , compare with the work done on m1 by the
weight of m1, Wwr1 E ?
A.
Wwr3 E > Wwr1E
If h is the distance by which m3 has moved: h=1m
r
W3 E does positive work as m3 moves down: Wwr 3 E = m3 gh
r
W1E does negative work as m1 moves up:
Wwr1 E = − m1 gh
We know that m1<m3
B.
Wwr3 E < Wwr1 E
C.
Wwr3 E = Wwr1 E
D.
Wwr3 E = −Wwr1 E
E.
Can' t tell. There is not enough information.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
15. At t1, what is the kinetic energy of the three masses (in Joules)?
A. 0
B. 9.8
Apply the work energy theorem
For the system m1+m2+m3,



 ⇒ KE f = Wnet = 9.8 J
= (m3 − m1 ) gh 


1
(m1 + m2 + m3 )v 2
2
KEi = 0
KE f =
Wnet = WWr3 E + WWr1 E
C. 19.6
D. 29.4
E. 39.2
16. At t=0s, the total mechanical energy of the system of the three masses
and strings is 0J, what is its total mechanical energy at t1 (in Joules)?
A. 0
All forces that do work are conservative (namely the weights of
m1 and m3).
B. 9.8
C. 39.2
D. -9.8
E. -39.2
17. What is the value of t1 (in seconds)?
A. 1
B. 1.2
at t1, m3 has moved by h=1m
h=
C.
D.
E.
1.5
1.8
2.1
1 2
at1 ⇒ t1 =
2
2h
= 1.24 s
a
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Questions 18 through 21 all refer to the same problem.
A bar is constructed from three equally sized pieces made of different
materials as shown below. It has a total mass of m=3kg. The points A, B
and C are located at the center of mass of each piece.
A
B
C
5N
10 N
The bar is initially at rest on a frictionless horizontal surface. When forces
of 5N and 10N are applied at the locations shown, the bar begins to rotate
counterclockwise about its center of mass.
18. What can you say about the location of the center of mass of the bar?
Choose the statement that is the most precise.
A. It is to the left of B
B. It is at B
C. It is to the right of B
Since the bar starts rotating counterclockwise from rest, the net
torque about the CM of the bar is counterclockwise (out of the
page). The distance between the 5N force and the CM must be at
least twice as large as the distance between the CM and the 10N
force.
D. It is at A
E. It is between A and B
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
19. What is the direction of the acceleration of the center of mass?
A.
The net force acting on the bar is directed down (sum of the 5N
and 10N forces)
B.
C.
D.
E. undefined since it is 0
20. What is the magnitude of the acceleration of the center of mass (in
m/s2)?
A.
B.
C.
D.
0
1.67
3.33
5
Use Newton's 2nd law:
10 + 5 = 3a ⇒ a = 5m / s 2
E.
15
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
21. In a second experiment, the forces applied on the bar are doubled (the
5N force becomes 10N and the 10N force becomes 20N). Nothing
else changes.
r
How would the acceleration aCM of the center of mass and the
r
angular acceleration α bar of the bar compare between the first and the
second experiments?
A.
B.
C.
D.
E.
r exp 2
r exp1
r exp 2
r exp1
aCM
= aCM
and α CM
= α CM
r exp 2
r exp1
r exp 2
r exp1
aCM
= 2 aCM
and α CM
= 2 α CM
The net force and the net torque increase by a factor of 2. Thus
both linear and angular accelerations increase by a factor of 2.
r exp 2
r exp1
r exp 2
r exp1
aCM
= 2 aCM
and α CM
= α CM
r exp 2
r exp1
r exp 2
r exp1
aCM
= aCM
and α CM
= 2 α CM
r exp 2 1 r exp1
aCM
= aCM
2
and
r
exp 2
α CM
=
1 r exp1
α CM
2
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Questions 22 through 27 all refer to the same problem.
On a frictionless horizontal table, one puck A of mass m moves with speed
2v to the right. An identical puck B moves with speed v to the left as
indicated on the figure below.
At t=0 they collide with a massless rod and stick to the ends of the rod.
Error!
m
y
A
2v
b
x
z (out of the
page)
massless
rod
v B
t<0
b/2
m
t=0
22. What is the linear momentum of the system rod +A +B at t<0?
A.
B.
2mvxˆ
mvxˆ
ptotal = 2mvxˆ − mvxˆ = mvxˆ
C. − mvxˆ
D. 0
E. 3mvxˆ
23. What is the linear momentum of the system rod+A+B at t>0?
A. 2mvxˆ
mvxˆ
B. The net force on the system A+B+rod is 0. The total linear
momentum is conserved.
C. − mvxˆ
D. 0
E. 3mvxˆ
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
24. What is the angular momentum of the system rod+A+B just before
the collision in the center of mass frame of the system rod+A+B?
( zˆ is a unit vector out of the page on the figures above)
r
Hint:
recall that for a point particle of mass m and velocity
v,
r r
r
v
r
L = r × mv and that for a system of particles Ltotal = ∑ L
A.
2bmvˆz
B.
3
bmvˆz
2
C. 0
D. − 2bmvˆz
E.
3
− bmvˆz
2
The velocity of the center of mass is
r
2mvxˆ − mvxˆ 1
vCM =
= mvxˆ
3m
3
r
1   b 
1 
3
b


Ltotal = yˆ × m 2vxˆ − vxˆ  +  − yˆ  × m − vxˆ − vxˆ  = − bmvzˆ
2
3   2 
3 
2


25. What is the angular momentum of the system rod+A+B just after the
collision in the center of mass frame of the system rod+A+B?
( zˆ is a unit vector out of the page on the figures above)
Hint: what is the net torque acting on the system rod + A + B?
A.
B.
2bmvˆz
3
bmvˆz
2
C. 0
D. − 2bmvˆz
E.
3
− bmvˆz
2
There is no net torque on the system rod + A + B. The angular
momentum of the system is conserved.
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
26. After the collision, what is the moment of inertia of the system
rod+A+B about an axis of rotation going through its center of mass
and parallel to zˆ ?
A.
B.
1 2
mb
4
1 2
mb
2
2
2
1
b
b
I = m  + m  = mb 2
2
2
2
C.
D.
E.
mb 2
2mb 2
4mb 2
27. After the collision, what is the magnitude ω of the angular velocity of
the system rod+A+B about its center of mass?
A. 0
B.
C.
D.
v
2b
2v
b
3v
b
L = Iω ⇒ ω =
E.
4v
b
3v
3bmv
=
1
b
2 mb 2
2
1/--pages
Пожаловаться на содержимое документа